Solve for $x$. $\dfrac{7^{7}}{7^3}=7^x$ $x=$
Solution: When powers have the same base, $\dfrac{x^m}{x^n}=x^{m-n}$. Let's expand the powers for $\dfrac{7^{7}}{7^3}=7^x$. $\begin{aligned} \dfrac{7^{7}}{7^3}&=\dfrac{\overbrace{\cancel 7\cdot \cancel 7\cdot \cancel 7\cdot 7\cdot 7\cdot 7\cdot 7}^\text{7 times}}{\underbrace{\cancel 7\cdot \cancel 7\cdot \cancel 7}_\text{3 times}} \\\\\\ &=\underbrace{7\cdot 7\cdot 7\cdot 7}_{x\text{ times}} \\\\ \end{aligned}$ $x = 4$